# 22° halo

## Strange rainbows

The other day after lunch, I was taking a walk around the outskirts of our institute with Sasaki-san, our super-handyman. Suddenly he discovered a funny, rainbow-like coloured arc in the sky: When we took a closer look at the sky, we found a beautiful halo around the sun: Please note that you can click all figures to see them in higher resolution.

We showed it to our colleagues who were also excited to see it.

The broad arc that we saw first was about twice as far from the sun as the full halo, and only appeared below the sun, not all the way around. After reading up on all the various types of optical phenomena that can be seen around the sun, I would say (without much confidence, though) that this was a circumhorizon arc. This particular kind of arc can only appear when the sun is high in the sky, which means only at latitudes south of Copenhagen. For that reason, this was probably the first time I saw such an arc.

The halo, on the other hand, I have seen several times before, around the sun as well as the moon. It is called a 22° halo and is apparently very common. Being optical physicists, we had to briefly discuss the mechanism behind its formation. None of us had any clear idea about it, except that I was pretty sure that it happens through some kind of refraction or reflection in ice crystals in the atmosphere. So I looked it up on Wikipedia and got a pretty clear idea about it, but decided to work out the details for myself as a little exercise. In the following I will describe some simple physics and calculations that answers these questions:

• Why does the halo appear at 22° separation from the sun?
• Why is the sky darker on the inside than on the outside of the halo?
• Why are the green and blue colours harder to see than the red and yellows — and why are the colours separated at all?

## 22° what?

The 22° halo gets its name from its separation from the sun. Perhaps it is not immediately obvious what 22° separation means, so let me get that settled first.

Here we see Sasaki-san observing the halo, although when I took this photo it had almost faded away, so it is a bit hard to see. He is a wise guy, and please notice how he blocks the light coming directly from the sun with his hand – a clever little trick I can warmly recommend when looking at the sky around the sun!1 The halo forms a nice circle centred on the sun. If we want to talk about the size of this circle, how should we define its radius? Since we don’t know how far away from us it is, it is also pretty hard to tell its radius in kilometres. That is why distances in the sky are usually measured in degrees of the angle between the two imaginary lines extending from our eyes to the two points of interest in the sky. The following diagram shows Sasaki-san, who — with his keen eye — can measure the distance of the halo to the sun to be quite close to 22°. In case you don’t fully trust Sasaki-san’s estimation, I have prepared a little evidence, based on the photo above of the full halo. That photo was taken on my Canon EOS 7D camera with a zoom lens set to 15 mm focal length. I cropped the photo in the sides but not at top and bottom. With a field-of view calculator we find that this camera-lens combination gives a field-of-view of 52.5° in the vertical (Y) direction. The field-of-view is the angular distance covered by the image.

In the following figure I have added two circles centred on the sun. The blue circle has a radius that is equal to half the image height, 1080 pixels, and therefore corresponds to 52.5°/2 = 26.3°. The red circle extends roughly to the inner edge of the halo. Its radius is 887 pixels, which in angular distance then becomes 26.3° $\times$ 887 px/1080 px = 21.6°. ## Ice in the sky

Okay, now we have established that this particular halo had a radius of 22°. The reason for its name is of course that it always appear at this same angle from the sun. So why is that?

First, let me skip very quickly past the actual atmospheric conditions that lead to the appearance of this phenomenon, since meteorology is not one of my strong sides. Basically, there are some thin cirrostratus clouds in the atmosphere consisting of ice crystals. Those crystals are hexagonal and can apparently be quite long, like a little stick: I grabbed this drawing from the Atmospheric Optics website, which has a more detailed explanation of how the shape and alignment of the ice crystals establishes the conditions necessary for the formation of the halo. Their description (over 3–4 pages) is great, so head over there after reading through my post — if I haven’t drained you for all interest in halos by then…

The essence of the story is that the hexagonal ice crystals are “poorly aligned”, which means that a small fraction of them have just the right angle to deflect the sunlight towards you, while others are aligned so as to direct the light towards another guy a few kilometres away from you. If the crystals were all aligned with each other, the halo would probably only be visible from a single location.

## Breaking the beams

So, getting to the interesting part (at least for me): Calculating the deflection angle of a ray of light through a single ice crystal. Yeah! Exciting, right?

You will see on the previous drawing that the light beam changes direction when passing through the crystal — once when entering, and again when exiting the crystal. This effect is called refraction and happens whenever light passes from one medium into another if the speed of light in the two media are different. It is exactly the same thing that happens when you look into a pool or bathtub and find your legs bent.

The speed of light in vacuum, c, is a constant of nature, but in other materials light usually travels slower than in vacuum. The ratio between c and the speed of light in a given material, v, is called the index of refraction, denoted by n:

For ice, the index of refraction is around 1.31. Light therefore travels 1.31 times slower in ice than in air (which has $n$ very close to 1). When light slows down, the wavelength also becomes shorter. Very loosely speaking, the wave of light then has to make a break upon entering the ice in order to keep itself together, as shown in the next figure. The amount of “breaking,” or refraction, is governed by Snell’s law, which relates the angles of the inbound and outbound beams:

$\theta_1$ and $\theta_2$ are the angles with the surface normal (the dotted line) of the inbound and outbound beams, respectively, while $n_1$ and $n_2$ are the refractive indices of the two materials. The formula says that the product of refractive index and the sine of the beam direction must be the same on either side of the interface.

Let us now use this formula to calculate exactly how much a beam of sunlight is deflected when going through a hexagonal crystal of ice.

## A bit of geometry

We assume that the ice crystals are perfectly hexagonal, that is, with 120° between each equal-length side face. A ray of light enters the crystal on one side at some angle $\theta_1$.2 If that angle is small, that is, if the ray enters almost perpendicularly to the surface, the ray will exit the crystal at the opposite face. In that case, there will be no deflection of the light and consequently that crystal will not contribute to any halo effect.

If the angle is instead somewhat larger, the ray will cross through the crystal and come out at the next-neighbouring crystal face. I illustrated that on the following figure, together with a bunch of guiding lines and angles. Remember again that you can click the drawing to see a larger version. The solid orange line is the path of the ray. The dashed blue lines are the surface normals (perpendicular lines). The thin dashed orange lines are parallel to the original beam from the sun. Proceeding step by step, we first find the refracted angle $\theta_2$ of the ray after entering the crystal using Snell’s law:

The angle between the two surface normals is 120°, and since the angle sum of a triangle is 180°, we find for $\theta_3$:

The refracted angle after the ray’s passage through the second ice-air interface is again obtained from Snell’s law:

After passage through the ice crystal, the ray of light has been deflected from its original direction by the angle denoted by $\theta_d$ on the drawing. To find that angle, I constructed an extra triangle on the right where one angle is 120°, another equal to $\theta_1$, and the last one therefore 60° – $\theta_1$.3 But that last angle is also the difference between $\theta_4$ and our unknown $\theta_d$, so finally we find for the deflection angle:

an expression now dependent only on the incidence angle $\theta_1$ and the known refractive indices $n_{\mathrm{air}}=1$ and $n_{\mathrm{ice}}=1.31$.

Now, let us plot these relations to see how the angle $\theta_4$ of the outgoing beam and the deflection of the sunbeam $\theta_d$ depends on the incidence angle $\theta_1$: First of all, we see from the red curve that the angle of the outgoing beam becomes larger when the incidence angle becomes smaller, that is, when the direction of the beam entering the crystal gets closer to perpendicular. When this incident direction gets as low as 13.5°, the direction of the outgoing beam will be 90° — parallel with the surface. For incidence angles lower than that, no light will exit the crystal at the surface we are interested in. All the light hitting that surface will be reflected instead of being refracted through the interface. This process is called total internal reflection and occurs for angles larger than the critical angle

In the case of the interface from ice into air, $n_1=n_{\mathrm{ice}}$ and $n_2=n_{\mathrm{air}}$, and the critical angle becomes $\theta_c$ = 49.8°. Indeed, for $\theta_1$ = 13.5° we get from the above equations that $\theta_3$ = 49.8°.

Next, let us look at the deflection angle (blue curve), which is the one that determines the size of the halo. It covers a wide range of angles up to about 45°, but mostly it is concentrated around the lower 20’s. The minimum value is 21.8°, consistent with what we estimated from the photograph above. It is interesting to note that this minimum deflection happens in the symmetric case where $\theta_1$ = $\theta_4$ = 41°. The minimum is marked by the grey lines.

From this plot, we can now answer two of the three questions I posed in the introduction:

Why does the halo appear at 22° separation from the sun?

A large part of the light that makes up the halo is deflected at an angle of 22° or slightly above. From the plot it seems like a lot of the light is also deflected at larger angles, but as mentioned in a footnote at Atmospheric Optics, a couple of factors combine to decrease the contribution to the halo from these larger angles. See also my list of extra questions at the end of the post.

We may now have convinced ourselves that the light from the sun is deflected at roughly 22° in the ice crystals in the upper atmosphere that are properly aligned. But, if we go back and look at the halo from Sasaki-san’s point of view, it may not be immediately obvious why he should see the halo at the same 22° angle of separation from the sun: The 22° degree deflection angle that we just found is the one marked by orange in the figure — it is not the same as the angle $\theta_h$ between the sun and the halo as seen by Sasaki-san. Indeed, if we study the triangle in the figure and remember that its angle sum should be 180°, we find that the halo angle should be

where $\theta_s$ is the angle between Sasaki-san and the halo as seen from the sun. But the figure is of course deceiving. The sun is much, much further away than it is drawn — roughly 150 million kilometres — while the ice crystals forming the halo are only a few kilometres away. For that reason, $\theta_s$ is essentially zero, and the size of the halo becomes identical to the deflection angle $\theta_d$.

Why is the sky darker on the inside than on the outside of the halo?

From the plot above, we see that light is deflected by the ice into a wide range of directions larger than 22°, but no light goes below that angle. This means that no light is added by the ice into the inside of the halo, at angles lower than 22°, but extra light that we would not have seen in the absence of the ice reaches our eyes from the parts of the sky on the outside of the halo. Inside the halo, the sky is just the same brightness as it would otherwise have been without the ice.

For the answer to the third question we have to take a quick look at another physical concept.

## Dispersion — why rainbows are so colourful

Apart from introducing another wonderful tidbit of physics, bringing up this subject also lets me show you another one of the photos I took of the halo that day: It is easier to see the rainbow-like colouring of the halo on this close-up photo (with our new office building in the front). So far in the discussion, I have not talked much about colour and its influence on the propagation of the light. Also, I claimed that the refractive index of ice is 1.31. This is not wrong, but also not the full story. There is a link between colour and refractive index: dispersion.

In most materials, the index of refraction increases for decreasing wavelength. Blue light (short wavelength) therefore travels slightly slower than red light. Here I plotted the dependence of the refractive index of ice on wavelength/colour:45 The refractive index is about 1% larger for violet than for red light. This means that the plot of the deflection angle $\theta_d$ from before should be modified – each colour gets its own curve. There is an interactive figure, equivalent to this plot, at Atmospheric Optics where you can rotate the ice crystal (change $\theta_1$) and see how it affects the deflection angle of red and blue light.

From this plot, it is now clear why the inner part of the halo is red, while the outer part becomes blueish. Sunlight consist of colours from all over the visible spectrum. Usually this mix of colours appears like white light with a slight yellow tint, but when the light hits a refracting object, like ice, water or a glass prism, it splits all its constituent wavelengths into separate directions.

In the case of our ice crystal, the minimum deflection angle for red is 21.5°, while for blue it is around 22.2°. At low angles in the halo, only red light gets deflected, while at larger angles the other colours of the spectrum start to gradually appear. The reason why the yellow, green and blue areas are less distinct than the red part is that red light also deflects into those outer areas of the halo. The total colour at larger angles are therefore a mix of the different colours, while the colours at small angles are purely red. For example, the part of the halo where the green light starts to appear will look more yellowish because of the mixing with red light.

## There’s more to the story than that

This blog post is already excessively long, so I think I should stop just about here. Hopefully I got through the explanation of the main principles involved in the making of a 22° halo. If I have made some errors or if you find some parts to be too difficult to understand, please let me know in the comments. I would also love to hear if you have seen this kind of halo around the sun or the moon. Do you see it often? The Wikipedia article claims that it may be seen as often as 100 days a year!

Finally, if you are not completely exhausted by now and simply can’t get enough of geometrical optics, I invite you to consider the following questions. These are questions I would be happy to study myself – but I guess I have already spent too many evenings on this halo. Do let me know if you actually go through these calculations!

### Questions

• The light hitting the crystal is of course not a simple ray, but rather a wide beam. For different crystal orientations, how large a part of the beam enters the crystal and exits from the right surface?
• Not all of the light that hits a surface is transmitted – a certain portion is reflected instead. The ratios of transmission and reflection are dependent on the light’s polarization, wavelength and incidence angle (recall from my description of total internal reflection that light entering above the critical angle is fully reflected). Exactly how much light is reflected at the ice crystal’s surface instead of refracted through for different angles and wavelengths? This can be calculated from the Fresnel equation.
• In these calculations, I assumed that light entered perpendicularly to the crystal axis. What if that is not the case?
• If the ice crystals are homogeneously aligned, how large a fraction of them will contribute to the halo?
• I only investigated the 22° halo. What is the optics behind the circumhorizon arc in the very first picture?

Feel free to base your calculations on the Python script I wrote to generate the plots in this post: halo-calculations.py

1. For several hours after lunch break I had irritated eyes and found it difficult to focus. I should have taken the danger of direct sunlight more seriously and looked at the halo for a shorter time. At least I made very sure to use the camera’s LCD screen instead of the viewfinder for photographing the sun. I hope the sensor was not damaged too much in the process…?

2. I also assume that the ray is entering perpendicularly to the crystal axis.

3. For $\theta_1 > 60°$ the construction is a little different, but the result remains the same.

4. Based on the formula on this Wikipedia page, originally published in Release on the Refractive Index of Ordinary Water Substance as a Function of Wavelength, Temperature and Pressure by IAWPS. I assume a temperature of -20°C and a density of ice of 920 kg/m³. See also WolframAlpha

5. The colour spectrum is based on code from the Color Science page.